Using Stirling's formula, show that limit is equal to zero
$begingroup$
Let $lambdain(0,1)$,
$h=lfloor(i+1)lambdarfloor $.
Show
$$
lim_{itoinfty}binom{i}{h}(1-lambda)^{(i-h)}lambda^{h}=0
$$
I was able to prove that for $lambda=frac{1}{2}$ using approximation for central binomial coefficent
$binom{2n}{n}approxfrac{4^n}{sqrt{pi n}}$. But not the other cases.
limits asymptotics binomial-coefficients
asked Jan 9 at 21:07
user121882user121882
6615
$endgroup$
add a comment |
$begingroup$
Let $lambdain(0,1)$,
$h=lfloor(i+1)lambdarfloor $.
Show
$$
lim_{itoinfty}binom{i}{h}(1-lambda)^{(i-h)}lambda^{h}=0
$$
I was able to prove that for $lambda=frac{1}{2}$ using approximation for central binomial coefficent
$binom{2n}{n}approxfrac{4^n}{sqrt{pi n}}$. But not the other cases.
limits asymptotics binomial-coefficients
asked Jan 9 at 21:07
user121882user121882
6615
$endgroup$
add a comment |
$begingroup$
Let $lambdain(0,1)$,
$h=lfloor(i+1)lambdarfloor $.
Show
$$
lim_{itoinfty}binom{i}{h}(1-lambda)^{(i-h)}lambda^{h}=0
$$
I was able to prove that for $lambda=frac{1}{2}$ using approximation for central binomial coefficent
$binom{2n}{n}approxfrac{4^n}{sqrt{pi n}}$. But not the other cases.
limits asymptotics binomial-coefficients
asked Jan 9 at 21:07
user121882user121882
6615
$endgroup$
Let $lambdain(0,1)$,
$h=lfloor(i+1)lambdarfloor $.
Show
$$
lim_{itoinfty}binom{i}{h}(1-lambda)^{(i-h)}lambda^{h}=0
$$
I was able to prove that for $lambda=frac{1}{2}$ using approximation for central binomial coefficent
$binom{2n}{n}approxfrac{4^n}{sqrt{pi n}}$. But not the other cases.
limits asymptotics binomial-coefficients
limits asymptotics binomial-coefficients
asked Jan 9 at 21:07
user121882user121882
6615
asked Jan 9 at 21:07
user121882user121882
6615
asked Jan 9 at 21:07
user121882user121882
6615
asked Jan 9 at 21:07
user121882user121882
6615
asked Jan 9 at 21:07
user121882user121882
6615
6615
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint: First note that $hto infty$ and $i-hto infty$ as $ito infty$.
Apply the Stirling approximation for the binomial
$$ binom{i}{h} =frac{i!}{h! (i-h)!}approx
sqrt frac{i}{2 pi h (h-i)} left(frac{i}{i-h}right)^i left(frac{i-h}{h}right)^h $$
And write $h=(i+1)lambda-epsilon$ with $0le epsilon <1$, replace and evaluate the limit.
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: First note that $hto infty$ and $i-hto infty$ as $ito infty$.
Apply the Stirling approximation for the binomial
$$ binom{i}{h} =frac{i!}{h! (i-h)!}approx
sqrt frac{i}{2 pi h (h-i)} left(frac{i}{i-h}right)^i left(frac{i-h}{h}right)^h $$
And write $h=(i+1)lambda-epsilon$ with $0le epsilon <1$, replace and evaluate the limit.
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
$endgroup$
add a comment |
$begingroup$
Hint: First note that $hto infty$ and $i-hto infty$ as $ito infty$.
Apply the Stirling approximation for the binomial
$$ binom{i}{h} =frac{i!}{h! (i-h)!}approx
sqrt frac{i}{2 pi h (h-i)} left(frac{i}{i-h}right)^i left(frac{i-h}{h}right)^h $$
And write $h=(i+1)lambda-epsilon$ with $0le epsilon <1$, replace and evaluate the limit.
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
$endgroup$
add a comment |
$begingroup$
Hint: First note that $hto infty$ and $i-hto infty$ as $ito infty$.
Apply the Stirling approximation for the binomial
$$ binom{i}{h} =frac{i!}{h! (i-h)!}approx
sqrt frac{i}{2 pi h (h-i)} left(frac{i}{i-h}right)^i left(frac{i-h}{h}right)^h $$
And write $h=(i+1)lambda-epsilon$ with $0le epsilon <1$, replace and evaluate the limit.
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
$endgroup$
Hint: First note that $hto infty$ and $i-hto infty$ as $ito infty$.
Apply the Stirling approximation for the binomial
$$ binom{i}{h} =frac{i!}{h! (i-h)!}approx
sqrt frac{i}{2 pi h (h-i)} left(frac{i}{i-h}right)^i left(frac{i-h}{h}right)^h $$
And write $h=(i+1)lambda-epsilon$ with $0le epsilon <1$, replace and evaluate the limit.
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
answered Jan 12 at 3:23
leonbloyleonbloy
40.6k645107
40.6k645107
add a comment |
add a comment |
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